AC Voltage Applied to a Series LCR Circuit

Alternating Current · CBSE

Theory

The circuit

Diagram in words

An inductor $L$ , a capacitor $C$ and a resistor $R$ joined end to end in one loop across an ac source $v = v_m\sin\omega t$ . Being in series, the same current flows through all three at every instant; only the voltages across them differ in phase.

Kirchhoff's loop rule gives the governing equation $L\dfrac{di}{dt} + iR + \dfrac{q}{C} = v_m\sin\omega t$ ; the phasor method gives its steady-state solution.

Phasor-diagram derivation (the standard 5-mark answer)

Let the current be $i = i_m\sin(\omega t + \phi)$ , the same in every element. Draw the current phasor $\mathbf I$ along a reference direction.
The three voltage phasors, with amplitudes $v_{Rm} = i_mR$ , $v_{Lm} = i_mX_L$ , $v_{Cm} = i_mX_C$ :
$\mathbf V_R$ is parallel to $\mathbf I$ (in phase with the current).
$\mathbf V_L$ is $90^\circ$ ahead of $\mathbf I$ (voltage leads current in an inductor).
$\mathbf V_C$ is $90^\circ$ behind $\mathbf I$ (voltage lags current in a capacitor).
$\mathbf V_L$ and $\mathbf V_C$ lie along the same line in opposite directions, so they combine into a single phasor of magnitude $\lvert v_{Lm} - v_{Cm}\rvert$ , perpendicular to $\mathbf V_R$ .
The source voltage phasor $\mathbf V$ is the resultant, the hypotenuse of a right triangle with sides $\mathbf V_R$ and $(\mathbf V_L + \mathbf V_C)$ :

$v_m^2 = v_{Rm}^2 + (v_{Lm} - v_{Cm})^2 = i_m^2\left[R^2 + (X_L - X_C)^2\right]$

$i_m = \dfrac{v_m}{\sqrt{R^2 + (X_L - X_C)^2}} = \dfrac{v_m}{Z}$

The impedance of the circuit is therefore

$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + \left(\omega L - \dfrac{1}{\omega C}\right)^2}$

SI unit of $Z$ : ohm ( $\Omega$ ). In rms terms, $I = V/Z$ and $V = \sqrt{V_R^2 + (V_L-V_C)^2}$ .

Phase angle and the impedance triangle

Since $\mathbf I$ is always parallel to $\mathbf V_R$ , the phase angle is the angle between $\mathbf V_R$ and $\mathbf V$ :

$\tan\phi = \dfrac{v_{Lm} - v_{Cm}}{v_{Rm}} = \dfrac{X_L - X_C}{R} = \dfrac{\omega L - 1/\omega C}{R}$

Impedance diagram: a right triangle with base $R$ , perpendicular $(X_L - X_C)$ and hypotenuse $Z$ , the angle at the base being $\phi$ . It also gives $\cos\phi = R/Z$ directly.
If $X_L > X_C$ the circuit is predominantly inductive and the current lags the source voltage. If $X_C > X_L$ it is predominantly capacitive and the current leads. If $X_L = X_C$ the two are in phase.
Sign convention: NCERT writes $i = i_m\sin(\omega t+\phi)$ with $\tan\phi = (X_C - X_L)/R$ (so a positive $\phi$ means the current leads). Board answers commonly write $\tan\phi = (X_L - X_C)/R$ for the angle by which the voltage leads the current. Both are accepted; always add the sentence saying which one leads.
The phasor solution is the steady-state solution; the full solution also contains a transient part that dies away.

The two-element cases CBSE sets constantly

Circuit	Impedance	Phase angle	Which leads
Series RL	$Z = \sqrt{R^2 + X_L^2}$	$\tan\phi = \dfrac{X_L}{R}$	voltage leads current
Series RC	$Z = \sqrt{R^2 + X_C^2}$	$\tan\phi = \dfrac{X_C}{R} = \dfrac{1}{\omega RC}$	current leads voltage
Series LC, $R\approx0$	$Z = \lvert X_L - X_C\rvert$	$\phi = \pm\dfrac{\pi}{2}$	wattless current, $P = 0$

Peak drops in a series RC circuit: $V_{0R} = \dfrac{V_0R}{\sqrt{R^2+X_C^2}}$ and $V_{0C} = \dfrac{V_0X_C}{\sqrt{R^2+X_C^2}}$ ; the same forms with $X_L$ apply to a series RL circuit.
With any resistance in the circuit, $\phi = \tan^{-1}(X/R)$ is strictly less than $\pi/2$ — a phase difference of exactly $\pi/2$ requires a purely reactive circuit.

Why the voltages do not add arithmetically

$V_R$ , $V_L$ and $V_C$ are not in phase with one another, so they cannot be added as ordinary numbers; they must be combined as phasors. Hence the rms source voltage is $V = \sqrt{V_R^2 + (V_L-V_C)^2}$ , generally less than $V_R + V_L + V_C$ .
Worked pattern (2023): $R = 30\ \Omega$ and $X_C = 40\ \Omega$ on a 200 V, 50 Hz supply give $Z = 50\ \Omega$ , $I = 4$ A, $V_R = 120$ V, $V_C = 160$ V. Their arithmetic sum is 280 V, more than the 200 V source — no paradox, because $V = \sqrt{120^2+160^2} = 200$ V, exactly the source voltage.
Same idea in one line for a meter question: if $V_1$ across the $L$ – $C$ combination reads 40 V on a 50 V rms source, then $V_2$ across $R$ is $\sqrt{50^2-40^2} = 30$ V.

Resonance

Definition: the series LCR circuit is at resonance when $X_L = X_C$ , so that the current amplitude is a maximum and the current is in phase with the applied voltage.

$\omega_0 L = \dfrac{1}{\omega_0 C} \;\Rightarrow\; \omega_0 = \dfrac{1}{\sqrt{LC}}, \qquad \nu_0 = \dfrac{1}{2\pi\sqrt{LC}}$

At resonance: $Z = R$ (its minimum value), $i_m = v_m/R$ (maximum), $\phi = 0$ , power factor $= 1$ , and the power dissipated $VI = V^2/R$ is a maximum.
$V_L$ and $V_C$ are then equal in magnitude but $180^\circ$ out of phase, so they cancel completely and the whole source voltage appears across $R$ . Each may individually be many times the source voltage: with $R = 20\ \Omega$ , $L = 2$ H, $C = 50\ \mu$ F on a 200 V supply, $\omega_0 = 100$ rad s $^{-1}$ , $I = 10$ A, and the drop across the capacitor is $I/\omega_0C = 2000$ V.
Resonance requires both $L$ and $C$ : only then can the two voltages cancel. An RL or RC circuit cannot resonate.
Graph of $Z$ against $\omega$ (or $\nu$ ): large at low frequency where $1/\omega C$ dominates, falling to the minimum $Z = R$ at $\omega_0$ , rising again at high frequency where $\omega L$ dominates — a V-shaped curve. Correspondingly the current first increases and then decreases as the frequency is raised.
Graph of current amplitude against $\omega$ for two resistances $R_1 > R_2$ : both peak at $\omega_0$ , but the $R_2$ curve is taller (peak $v_m/R_2$ ) and narrower; far from resonance the two curves nearly coincide.
Applications: tuning a radio (the capacitance is varied until the circuit's resonant frequency matches the station's, so that station's current amplitude becomes maximum) and the metal detector at an airport (a metal object changes the effective inductance and hence the impedance of a resonant coil, and the change in current triggers the alarm).

Sharpness of resonance and the quality factor

Sharpness of resonance is measured by the quality factor $Q$ , defined as the ratio of the voltage developed across $L$ (or across $C$ ) at resonance to the voltage developed across $R$ :

$Q = \dfrac{\omega_0 L}{R} = \dfrac{1}{\omega_0 CR} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}$

Equivalently $Q = \dfrac{\omega_0}{2\Delta\omega}$ , the ratio of the resonant angular frequency to the bandwidth $2\Delta\omega = R/L$ (the frequency width over which the current stays above $1/\sqrt2$ of its peak).
$Q$ is a ratio of like quantities: it is dimensionless and has no SI unit.
Large $L$ and small $R$ give large $Q$ , a taller and sharper resonance curve, and a more selective circuit — which is exactly what makes tuning precise. Among a family of resonance curves, the tallest and sharpest belongs to the smallest resistance.

Standard results and worked patterns

$Z = \sqrt{R^2+(X_L-X_C)^2}$ ; $I = V/Z$ ; $\tan\phi = (X_L-X_C)/R$ ; $V = \sqrt{V_R^2+(V_L-V_C)^2}$ .
$\dfrac{X_L}{X_C} = \omega^2 LC$ .
Resonance: $\omega_0 = 1/\sqrt{LC}$ , $\nu_0 = 1/(2\pi\sqrt{LC})$ , $C = \dfrac{1}{4\pi^2\nu_0^2L}$ , $L = \dfrac{1}{4\pi^2\nu_0^2C}$ .
Reading a source equation: $V = 40\sin(1000t + \pi/3)$ gives $V_0 = 40$ V and $\omega = 1000$ rad s $^{-1}$ ; with $L = 0.4$ H, $C = 10\ \mu$ F, $R = 400\ \Omega$ : $X_L = 400\ \Omega$ , $X_C = 100\ \Omega$ , $Z = 500\ \Omega$ , $I_0 = 0.08$ A.
Phase-angle pattern: $\omega = 1000$ rad s $^{-1}$ , $L = 100$ mH, $C = 2\ \mu$ F, $R = 400\ \Omega$ give $X_L = 100\ \Omega$ , $X_C = 500\ \Omega$ , so $\phi = \tan^{-1}\left(\dfrac{500-100}{400}\right) = 45^\circ$ with the current leading (capacitive).
Coil on dc then on ac: $R = V_{dc}/I_{dc}$ , $Z = V_{ac}/I_{ac}$ , $X_L = \sqrt{Z^2-R^2}$ , $L = X_L/2\pi\nu$ . On dc only $R$ limits the current; on ac the reactance adds in quadrature, so $Z > R$ and less current flows.
Brightness questions: the current, and hence the brightness of a series bulb, is $I = V/Z$ . Raising $C$ (dielectric slab) lowers $X_C$ ; raising $L$ (iron rod, more turns) raises $X_L$ ; adding a capacitor with $X_C = X_L$ brings the circuit to resonance, making $Z = R$ minimum and the bulb brightest.
Detuning from resonance: if a resonant circuit has its capacitance increased (a dielectric inserted), $X_C$ falls below $X_L$ , $\tan\phi = (X_L-X_C)/R > 0$ , and the current now lags the voltage. For $\omega < \omega_0$ , $X_C > X_L$ and the current leads.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
What is the value of impedance of a resonant series LCR circuit?
1 markCBSE Class 12 Physics — Very Short Answer · 2020, R · Section A, 1 markAnswer 1 ↓
2
A series combination of an inductor ( $L$ ), a capacitor ( $C$ ) and a resistor ( $R$ ) is connected across an ac source of emf of peak value $E_0$ and angular frequency ( $\omega$ ). Plot a graph to show the variation of impedance of the circuit with angular frequency ( $\omega$ ).
1 markCBSE Class 12 Physics — Very Short Answer · 2020 · Section A, 1 markAnswer 2 ↓
3
An inductor $L=0.4$ H, a capacitor $C=10\ \mu\text{F}$ and a resistor $R=400\ \Omega$ are connected in series to an ac source $V=40\sin(1000\,t+\pi/3)$ volt. Calculate the (a) impedance of the circuit, and (b) peak value of current.
2 marksCBSE Class 12 Physics — SA · 2020C · Section B, 2 marksAnswer 3 ↓
4
Explain the term 'sharpness of resonance' in an ac circuit. (1/2, 2020) OR Define 'quality factor' of resonance in a series LCR circuit. What is its S.I. unit? (Delhi 2016)
2 marksCBSE Class 12 Physics — SA · Section B, 2 marksAnswer 4 ↓
5
A capacitor 'C', a variable resistor 'R' and a bulb 'B' are connected in series to the ac mains in the circuit shown (capacitor $C$ , in series with variable resistor $R$ , in series with bulb $B$ , connected across the ac mains). The bulb glows with some brightness. How will the glow of the bulb change if (i) a dielectric slab is introduced between the plates of the capacitor, keeping resistance $R$ the same; (ii) the resistance $R$ is increased, keeping the same capacitance?
2 marksCBSE Class 12 Physics — SA · Delhi 2014 · Section B, 2 marksAnswer 5 ↓
6
The figure shows a series LCR circuit connected to a variable-frequency $200$ V source with $L=50$ mH, $C=80\ \mu\text{F}$ and $R=40\ \Omega$ . Determine: (i) the source frequency which drives the circuit into resonance; (ii) the quality factor ( $Q$ ) of the circuit.
2 marksCBSE Class 12 Physics — SA · AI 2014C, An · Section B, 2 marksAnswer 6 ↓
7
A resistor of $30\ \Omega$ and a capacitor of $\dfrac{250}{\pi}\ \mu\text{F}$ are connected in series to a $200$ V, $50$ Hz ac source. Calculate (i) the current in the circuit, and (ii) the voltage drops across the resistor and the capacitor. (iii) Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox.
3 marksCBSE Class 12 Physics — SA · 2023 · Section C, 3 marksAnswer 7 ↓
8
A series LCR circuit with $R=20\ \Omega$ , $L=2$ H and $C=50\ \mu\text{F}$ is connected to a $200$ volts ac source of variable frequency. What is (i) the amplitude of the current, and (ii) the average power transferred to the circuit in one complete cycle, at resonance? (iii) Calculate the potential drop across the capacitor.
3 marksCBSE Class 12 Physics — SA · 2023 · Section C, 3 marksAnswer 8 ↓
9
(i) In an LCR series circuit connected to an ac source, the voltage and the current are in the same phase. If the capacitor is filled with a dielectric, will the current lead or lag behind or remain in phase with the voltage? Explain. (ii) In the circuit, why is the rms value of the net voltage not equal to the sum of the voltage drops across the individual elements? (iii) Draw a graph showing the variation of the impedance of the circuit with the frequency of the applied voltage.
3 marksCBSE Class 12 Physics — SA · 2020C · Section C, 3 marksAnswer 9 ↓
10
A resistance $R$ and a capacitor $C$ are connected in series to a source $V=V_0\sin\omega t$ . Find: (a) the peak value of the voltage across the (i) resistance and (ii) capacitor; (b) the phase difference between the applied voltage and current. Which of them is ahead?
3 marksCBSE Class 12 Physics — SA · 2020 · Section C, 3 marksAnswer 10 ↓
11
A resistor $R$ and an inductor $L$ are connected in series to a source $V=V_0\sin\omega t$ . Find the (a) peak value of the voltage drops across $R$ and across $L$ ; (b) phase difference between the applied voltage and the current. Which of them is ahead?
3 marksCBSE Class 12 Physics — SA · 2020, An · Section C, 3 marksAnswer 11 ↓
12
Find the value of the phase difference between the current and the voltage in the series LCR circuit shown below. Which one leads in phase: current or voltage? Given: $V=V_0\sin(1000t+\phi)$ , with $L=100$ mH, $C=2\ \mu\text{F}$ and $R=400\ \Omega$ all in series. (2/3, Delhi 2017) OR Determine the value of the phase difference between the current and the voltage in the given series LCR circuit ( $R=400\ \Omega$ , $C=2\ \mu\text{F}$ , $L=100$ mH, source $V=V_0\sin(1000t+\phi)$ ). (2/3, AI 2015)
3 marksCBSE Class 12 Physics — SA · Section C, 3 marksAnswer 12 ↓
13
An inductor $L$ of inductive reactance $X_L$ is connected in series with a bulb $B$ and an ac source. How would the brightness of the bulb change when (i) the number of turns in the inductor is reduced, (ii) an iron rod is inserted in the inductor, and (iii) a capacitor of reactance $X_C=X_L$ is inserted in series in the circuit? Justify your answer in each case.
3 marksCBSE Class 12 Physics — SA · Delhi 2015 · Section C, 3 marksAnswer 13 ↓
14
A source of ac voltage $V=V_0\sin\omega t$ is connected to a series combination of a resistor 'R' and a capacitor 'C'. Draw the phasor diagram and use it to obtain the expression for (i) the impedance of the circuit and (ii) the phase angle.
3 marksCBSE Class 12 Physics — SA · AI 2015C, Ev · Section C, 3 marksAnswer 14 ↓
15
The variation of inductive reactance ( $X_L$ ) of an inductor with the frequency ( $f$ ) of the ac source of $100$ V and variable frequency is shown in the figure: $X_L$ (in $\Omega$ ) versus $f$ (in Hz) is a straight line through the origin, passing through $(100\text{ Hz},\,20\ \Omega)$ (and hence also through $(200,40)$ and $(300,60)$ ). (i) Calculate the self-inductance of the inductor. (ii) When this inductor is used in series with a capacitor of unknown value and a resistor of $10\ \Omega$ at $300\ \text{s}^{-1}$ , maximum power dissipation occurs in the circuit. Calculate the capacitance of the capacitor.
5 marksCBSE Class 12 Physics — Long Answer · 2/5, 2020, Cr · Section E, 5 marksAnswer 15 ↓
16
(a) In a series LCR circuit connected across an ac source of variable frequency, obtain the expression for its impedance and draw a plot showing its variation with frequency of the ac source. (b) What is the phase difference between the voltages across the inductor and the capacitor at resonance in the LCR circuit? (c) When an inductor is connected to a $200$ V dc voltage, a current of $1$ A flows through it. When the same inductor is connected to a $200$ V, $50$ Hz ac source, only $0.5$ A current flows. Explain why. Also, calculate the self-inductance of the inductor.
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2019 · Section E, 5 marksAnswer 16 ↓
17
(a) What do you understand by 'sharpness of resonance' for a series LCR resonant circuit? How is it related to the quality factor 'Q' of the circuit? Using the graphs given in the figure (three peaked current-vs- $\omega$ curves of differing sharpness, labelled with resistances $R_1$ , $R_2$ , $R_3$ ), explain the factors which affect it. Which of the three curves — the sharpest and tallest, or the flattest and broadest — is the one with the minimum resistance ( $R$ )? (b) A $2\ \mu\text{F}$ capacitor, $100\ \Omega$ resistor and $8$ H inductor are connected in series with an ac source. Find the frequency of the ac source for which the current drawn in the circuit is maximum. If the peak value of emf of the source is $200$ V, calculate the (i) maximum current and (ii) inductive and capacitive reactance of the circuit at resonance.
5 marksCBSE Class 12 Physics — Long Answer · AI 2019, Cr · Section E, 5 marksAnswer 17 ↓
18
An ac source of voltage $V=V_0\sin\omega t$ is connected to a series combination of $L$ , $C$ and $R$ . Use the phasor diagram to obtain expressions for the impedance of the circuit and the phase angle between voltage and current. Find the condition when the current will be in phase with the voltage. What is the circuit called in this condition?
5 marksCBSE Class 12 Physics — Long Answer · 3/5, Delhi 2016, Ev · Section E, 5 marksAnswer 18 ↓
19
A $2\ \mu\text{F}$ capacitor, $100\ \Omega$ resistor and $8$ H inductor are connected in series with an ac source. (i) What should be the frequency of the source such that the current drawn in the circuit is maximum? What is this frequency called? (ii) If the peak value of emf of the source is $200$ V, find the maximum current. (iii) Draw a graph showing the variation of amplitude of circuit current with changing frequency of applied voltage in a series LCR circuit for two different values of resistance $R_1$ and $R_2$ ( $R_1>R_2$ ). (iv) Define the term 'Sharpness of Resonance'. Under what condition does a circuit become more selective?
5 marksCBSE Class 12 Physics — Long Answer · Foreign 2016, An · Section E, 5 marksAnswer 19 ↓
20
(a) A series LCR circuit is connected to an ac source of variable frequency. Draw a suitable phasor diagram to deduce the expressions for the amplitude of the current and the phase angle. (b) Obtain the condition of resonance. Draw a plot showing the variation of current with the frequency of the ac source for two resistances $R_1$ and $R_2$ ( $R_1>R_2$ ). Hence define the quality factor, $Q$ , and write its role in the tuning of the circuit.
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2014C, Ap · Section E, 5 marksAnswer 20 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

21
Which of the following statements about a series LCR circuit connected to an ac source is correct?
- (a)
  If the frequency of the source is increased, the impedance of the circuit first decreases and then increases.
- (b)
  If the net reactance ( $X_L-X_C$ ) of the circuit becomes equal to its resistance, the current leads the voltage by $45°$ .
- (c)
  At resonance, the voltage drop across the inductor is more than that across the capacitor.
- (d)
  At resonance, the voltage drop across the capacitor is more than that across the inductor.
1 markCBSE Class 12 Physics — Multiple Choice · 2023 · Section A, 1 markAnswer 21 ↓
22
What is the ratio of inductive and capacitive reactance in an ac circuit?
- (a)
  $\omega^2LC$
- (b)
  $LC^2$
- (c)
  $\dfrac{LC}{\omega^2}$
- (d)
  $\omega^2L$
1 markCBSE Class 12 Physics — Multiple Choice · 2023 · Section A, 1 markAnswer 22 ↓
23
In a circuit, the phase difference between the alternating current and the source voltage is $\dfrac{\pi}{2}$ . Which of the following cannot be the element(s) of the circuit?
- (a)
  only $C$
- (b)
  only $L$
- (c)
  $L$ and $R$
- (d)
  $L$ or $C$
1 markCBSE Class 12 Physics — Multiple Choice · Term I 2021-22 · Section A, 1 markAnswer 23 ↓
24
The impedance of a series LCR circuit is:
- (a)
  $R+X_L+X_C$
- (b)
  $\sqrt{\dfrac{1}{X_C^2}+\dfrac{1}{X_L^2}+R^2}$
- (c)
  $\sqrt{X_L^2-X_C^2+R^2}$
- (d)
  $\sqrt{R^2+(X_L-X_C)^2}$
1 markCBSE Class 12 Physics — Multiple Choice · Term I 2021-22 · Section A, 1 markAnswer 24 ↓
25
The voltage across a resistor, an inductor, and a capacitor connected in series to an ac source are 20 V, 15 V and 30 V respectively. The resultant voltage in the circuit is:
- (a)
  5 V
- (b)
  20 V
- (c)
  25 V
- (d)
  65 V
1 markCBSE Class 12 Physics — Multiple Choice · Term I 2021-22 · Section A, 1 markAnswer 25 ↓
26
A circuit is connected to an ac source of variable frequency. As the frequency of the source is increased, the current first increases and then decreases. Which of the following combinations of elements is likely to comprise the circuit?
- (a)
  $L$ , $C$ and $R$
- (b)
  $L$ and $C$
- (c)
  $L$ and $R$
- (d)
  $R$ and $C$
1 markCBSE Class 12 Physics — Multiple Choice · Term I 2021-22 · Section A, 1 markAnswer 26 ↓
27
A $15\ \Omega$ resistor, an $80$ mH inductor and a capacitor of capacitance $C$ are connected in series with a $50$ Hz ac source. If the source voltage and current in the circuit are in phase, then the value of capacitance is:
- (a)
  $100\ \mu\text{F}$
- (b)
  $127\ \mu\text{F}$
- (c)
  $142\ \mu\text{F}$
- (d)
  $160\ \mu\text{F}$
1 markCBSE Class 12 Physics — Multiple Choice · Term I 2021-22 · Section A, 1 markAnswer 27 ↓
28
A $300\ \Omega$ resistor and a capacitor of $\left(\dfrac{25}{\pi}\right)\mu\text{F}$ are connected in series to a $200$ V $-$ $50$ Hz ac source. The current in the circuit is:
- (a)
  0.1 A
- (b)
  0.4 A
- (c)
  0.6 A
- (d)
  0.8 A
1 markCBSE Class 12 Physics — Multiple Choice · Term I 2021-22 · Section A, 1 markAnswer 28 ↓
29
The selectivity of a series LCR a.c. circuit is large, when:
- (a)
  $L$ is large and $R$ is large
- (b)
  $L$ is small and $R$ is small
- (c)
  $L$ is large and $R$ is small
- (d)
  $L=R$
1 markCBSE Class 12 Physics — Multiple Choice · 2020 · Section A, 1 markAnswer 29 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
At resonance, $X_L=X_C$ , so $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{R^2+0}=R$ Thus the impedance of a resonant series LCR circuit equals its resistance $R$ — the minimum value the impedance can take.
2.↑ back to the question
$Z=\sqrt{R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2}$ The graph of $Z$ versus $\omega$ is V-shaped: large for small $\omega$ (dominated by $1/\omega C$ ), falling to a minimum $Z_{min}=R$ exactly at the resonant angular frequency $\omega_0=1/\sqrt{LC}$ , then rising again for $\omega>\omega_0$ (dominated by $\omega L$ ).
3.↑ back to the question
Comparing with $V=V_0\sin(\omega t+\phi)$ : $V_0=40$ V, $\omega=1000$ rad/s. $X_L=\omega L=1000\times0.4=400\ \Omega$ $X_C=\dfrac{1}{\omega C}=\dfrac{1}{1000\times10\times10^{-6}}=100\ \Omega$ (a) $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{400^2+(400-100)^2}=\sqrt{400^2+300^2}=\sqrt{250000}=500\ \Omega$ (b) $I_0=\dfrac{V_0}{Z}=\dfrac{40}{500}=0.08\text{ A}=80\text{ mA}$
4.↑ back to the question
Sharpness of resonance: it is defined as the ratio of the voltage developed across the inductance ( $L$ ) or the capacitance ( $C$ ) at resonance to the voltage developed across the resistance ( $R$ ): $Q=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$ It may also be defined as the ratio of the resonant angular frequency to the bandwidth of the circuit: $Q=\dfrac{\omega_r}{2\Delta\omega}$ . A sharper (higher- $Q$ ) resonance curve means the circuit is more selective. This same quantity $Q$ is called the quality factor of the resonant LCR circuit. Being a ratio of two like (voltage, or frequency) quantities, it is dimensionless and has no S.I. unit.
5.↑ back to the question
Impedance $Z=\sqrt{R^2+X_C^2}$ , and current $I=\dfrac{V}{Z}$ . (i) Introducing a dielectric increases the capacitance $C$ , so $X_C=\dfrac{1}{\omega C}$ decreases; hence $Z$ decreases and the current increases — the bulb glows brighter. (ii) Increasing $R$ (with $C$ , and hence $X_C$ , unchanged) increases $Z=\sqrt{R^2+X_C^2}$ , decreasing the current — the bulb glows dimmer.
6.↑ back to the question
(i) $\omega_0=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{50\times10^{-3}\times80\times10^{-6}}}=\dfrac{1}{\sqrt{4\times10^{-6}}}=\dfrac{10^3}{2}=500\text{ rad/s}$ $\nu_0=\dfrac{\omega_0}{2\pi}=\dfrac{500}{2\pi}\approx80\text{ Hz}$ (ii) $Q=\dfrac1R\sqrt{\dfrac{L}{C}}=\dfrac{1}{40}\sqrt{\dfrac{50\times10^{-3}}{80\times10^{-6}}}=\dfrac{1}{40}\sqrt{625}=\dfrac{25}{40}=0.625$
7.↑ back to the question
$X_C=\dfrac{1}{2\pi\nu C}=\dfrac{1}{2\pi\times50\times\dfrac{250}{\pi}\times10^{-6}}=\dfrac{10^6}{2\times50\times250}=40\ \Omega$ $Z=\sqrt{R^2+X_C^2}=\sqrt{30^2+40^2}=\sqrt{2500}=50\ \Omega$ (i) $I=\dfrac{V}{Z}=\dfrac{200}{50}=4\text{ A}$ (ii) Voltage across resistor $=IR=4\times30=120$ V; voltage across capacitor $=IX_C=4\times40=160$ V (iii) Algebraic sum $=120+160=280$ V, which exceeds the source voltage of $200$ V. There is no real paradox: $V_R$ and $V_C$ are $90°$ out of phase, so they cannot be added algebraically — they must be combined as phasors: $V=\sqrt{V_R^2+V_C^2}=\sqrt{120^2+160^2}=\sqrt{40000}=200$ V, which correctly reproduces the given source voltage.
8.↑ back to the question
At resonance, $\omega=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{2\times50\times10^{-6}}}=100\text{ rad/s}$ , and $Z=R=20\ \Omega$ . (i) $I=\dfrac{V}{Z}=\dfrac{200}{20}=10\text{ A}$ (ii) Average power, $P=VI=200\times10=2000\text{ W}$ (iii) Potential drop across capacitor $=I\times\dfrac{1}{\omega C}=10\times\dfrac{1}{100\times50\times10^{-6}}=2000\text{ V}$
9.↑ back to the question
(i) 'Voltage and current in phase' means the circuit is at resonance, $X_L=X_C$ . Filling the capacitor with a dielectric increases $C$ , so $X_C=\dfrac{1}{\omega C}$ decreases, making $X_C<X_L$ ; now $\tan\phi=\dfrac{X_L-X_C}{R}>0$ , so the current lags behind the voltage. (ii) $V_R$ , $V_L$ and $V_C$ are not in phase with each other ( $V_R$ in phase with $I$ , $V_L$ leading $I$ by $90°$ , $V_C$ lagging $I$ by $90°$ ), so they cannot be added algebraically — they must be combined as phasors: $V=\sqrt{V_R^2+(V_L-V_C)^2}$ , which is generally less than the arithmetic sum $V_R+V_L+V_C$ . (iii) $Z=\sqrt{R^2+\left(\omega L-\dfrac1{\omega C}\right)^2}$ : the graph of $Z$ vs $f$ is large at low frequency, decreases to a minimum $Z_{min}=R$ at the resonant frequency $f_0=\dfrac{1}{2\pi\sqrt{LC}}$ , then increases again at higher frequency (a V-shaped curve).
10.↑ back to the question
$Z=\sqrt{R^2+X_C^2}$ , peak current $I_0=\dfrac{V_0}{Z}=\dfrac{V_0}{\sqrt{R^2+X_C^2}}$ (a)(i) $V_{0R}=I_0R=\dfrac{V_0R}{\sqrt{R^2+X_C^2}}$ (a)(ii) $V_{0C}=I_0X_C=\dfrac{V_0X_C}{\sqrt{R^2+X_C^2}}$ (b) $\phi=\tan^{-1}\left(\dfrac{X_C}{R}\right)=\tan^{-1}\left(\dfrac{1}{\omega RC}\right)$ . In a series RC circuit, the current leads the applied voltage in phase.
11.↑ back to the question
$Z=\sqrt{R^2+X_L^2}$ , $I_0=\dfrac{V_0}{Z}$ (a) $V_{0R}=I_0R=\dfrac{V_0R}{\sqrt{R^2+X_L^2}}$ ; $V_{0L}=I_0X_L=\dfrac{V_0X_L}{\sqrt{R^2+X_L^2}}$ (b) $\phi=\tan^{-1}\left(\dfrac{X_L}{R}\right)$ . In an R–L circuit, the applied voltage leads the current in phase by this angle $\phi$ (equivalently, the current lags the voltage).
12.↑ back to the question
Comparing $V=V_0\sin(\omega t+\phi)$ with the given equation, $\omega=1000$ rad/s. $X_L=\omega L=1000\times100\times10^{-3}=100\ \Omega$ $X_C=\dfrac{1}{\omega C}=\dfrac{1}{1000\times2\times10^{-6}}=500\ \Omega$ $\phi=\tan^{-1}\left(\dfrac{X_C-X_L}{R}\right)=\tan^{-1}\left(\dfrac{500-100}{400}\right)=\tan^{-1}(1)=45°$ Since $X_C>X_L$ , the circuit is capacitive and the current leads the voltage in phase by $45°$ .
13.↑ back to the question
Impedance $Z=\sqrt{X_L^2+R^2}$ , current $I=\dfrac{V}{Z}$ . (i) Reducing the number of turns decreases $L$ , hence $X_L$ decreases; $Z$ decreases and current increases — the bulb glows brighter. (ii) Inserting an iron rod increases $L$ (high permeability), so $X_L$ and $Z$ increase; the current decreases — the bulb becomes dimmer. (iii) With $X_C=X_L$ added in series, $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{R^2+0}=R$ — this is the resonance condition, impedance is minimum, current is maximum — the bulb glows at its brightest.
14.↑ back to the question
Let $I=I_0\sin(\omega t+\phi)$ be the current. $V_R=IR$ is in phase with $I$ ; $V_C=IX_C$ lags $I$ by $90°$ . Phasor diagram: draw $V_R$ along the current direction and $V_C$ perpendicular to it (lagging by $90°$ ); their resultant (right-triangle sum) gives the applied voltage: $V=\sqrt{V_R^2+V_C^2}$ . (i) $V=I\sqrt{R^2+X_C^2}=I\sqrt{R^2+\dfrac{1}{\omega^2C^2}}$ , so $I=\dfrac{V}{Z}$ with impedance $Z=\sqrt{R^2+\dfrac{1}{\omega^2C^2}}$ (ii) $\tan\phi=\dfrac{V_C}{V_R}=\dfrac{X_C}{R}=\dfrac{1}{\omega RC} \Rightarrow \phi=\tan^{-1}\left(\dfrac{1}{\omega RC}\right)$ , with the current leading the voltage by this angle $\phi$ .
15.↑ back to the question
(i) From the graph, at $f=100$ Hz, $X_L=20\ \Omega$ . $X_L=2\pi fL \Rightarrow L=\dfrac{X_L}{2\pi f}=\dfrac{20}{2\times3.14\times100}=0.0318\text{ H}=31.8\text{ mH}$ (ii) Maximum power dissipation occurs at resonance ( $X_L=X_C$ ) at $f=300$ Hz ( $300\ \text{s}^{-1}$ ). From the graph's linear relation, $X_L=2\pi fL=2\times3.14\times300\times0.0318\approx60\ \Omega$ at this frequency. Setting $X_C=X_L$ : $\dfrac{1}{2\pi fC}=60 \Rightarrow C=\dfrac{1}{2\times3.14\times300\times60}\approx8.8\times10^{-6}\text{ F}=8.8\ \mu\text{F}$
16.↑ back to the question
(a) With $I$ the circuit current, $V_L=IX_L$ leads $I$ by $\pi/2$ , $V_C=IX_C$ lags $I$ by $\pi/2$ , $V_R=IR$ is in phase with $I$ . The net voltage is $E=\sqrt{V_R^2+(V_L-V_C)^2}=I\sqrt{R^2+(X_L-X_C)^2}=IZ$ , where $Z=\sqrt{R^2+\left(\omega L-\dfrac1{\omega C}\right)^2}$ The $Z$ -versus-frequency graph is large at both very low and very high frequency, dipping to a minimum $Z=R$ at the resonant frequency $f_0=\dfrac1{2\pi\sqrt{LC}}$ (a V-shaped curve). (b) At resonance $X_L=X_C$ , so $V_L$ and $V_C$ are equal in magnitude; but $V_L$ leads the current by $90°$ while $V_C$ lags the current by $90°$ , so $V_L$ and $V_C$ are exactly opposite in phase — the phase difference between them is $180°$ ( $\pi$ rad), which is why they cancel and the net (applied) voltage equals $V_R$ alone at resonance. (c) With dc, only the resistance $R$ of the coil limits the current: $R=\dfrac{V}{I}=\dfrac{200}{1}=200\ \Omega$ . With ac, the inductive reactance $X_L$ also limits the current, raising the impedance to $Z=\sqrt{R^2+X_L^2}>R$ , so a smaller current (0.5 A) flows. $Z=\dfrac{V}{I}=\dfrac{200}{0.5}=400\ \Omega$ $X_L=\sqrt{Z^2-R^2}=\sqrt{400^2-200^2}=\sqrt{120000}\approx346.4\ \Omega$ $L=\dfrac{X_L}{2\pi f}=\dfrac{346.4}{2\pi\times50}\approx1.1\text{ H}$
17.↑ back to the question
(a) Sharpness of resonance is the ratio of the voltage developed across $L$ (or $C$ ) at resonance to the voltage developed across $R$ : $Q=\dfrac1R\sqrt{\dfrac LC}$ ; equivalently $Q=\dfrac{\omega_r}{2\Delta\omega}$ (ratio of resonant angular frequency to bandwidth). At resonance the current amplitude is maximum, $i_m=V_m/R$ ; a smaller $R$ gives a larger, narrower (sharper) resonance peak, i.e. higher $Q$ and better tuning/selectivity — so the sharpest, tallest curve corresponds to the smallest resistance. So among the three curves, the TALLEST AND SHARPEST peak is the one with the minimum resistance $R$ .
[Note on this record: the original figure was lost in import, so the labels $R_1$ , $R_2$ , $R_3$ can no longer be matched to particular curves. The question now asks which curve — sharpest or flattest — carries the least resistance, which is the physics the original part was testing and is exactly what this solution establishes.] (b) The current is maximum at resonance: $\nu=\dfrac{1}{2\pi\sqrt{LC}}=\dfrac{1}{2\times3.14\times\sqrt{8\times2\times10^{-6}}}\approx39.8\text{ Hz}$ (i) At resonance $Z=R$ , so $I_0=\dfrac{E_0}{R}=\dfrac{200}{100}=2\text{ A}$ (ii) $X_L=2\pi\nu L=2\times3.14\times39.8\times8\approx2000\ \Omega$ ; $X_C=\dfrac{1}{2\pi\nu C}=\dfrac{1}{2\times3.14\times39.8\times2\times10^{-6}}\approx2000\ \Omega$ (equal, as required at resonance).
18.↑ back to the question
With current $I$ , $V_L=IX_L$ leads $I$ by $\pi/2$ ; $V_C=IX_C$ lags $I$ by $\pi/2$ ; $V_R=IR$ is in phase with $I$ . $V_L$ and $V_C$ are anti-parallel phasors, so their resultant is $(V_L-V_C)$ , perpendicular to $V_R$ . By Pythagoras, $V=\sqrt{V_R^2+(V_L-V_C)^2}=I\sqrt{R^2+(X_L-X_C)^2}$ $Z=\dfrac{V}{I}=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{R^2+\left(\omega L-\dfrac1{\omega C}\right)^2}$ $\tan\phi=\dfrac{X_L-X_C}{R}=\dfrac{\omega L-1/\omega C}{R}$ Current is in phase with voltage ( $\phi=0$ ) when $X_L=X_C$ , i.e. $\omega=\dfrac{1}{\sqrt{LC}}$ . This condition is called resonance, and the circuit is then called a resonant (acceptor) circuit.
19.↑ back to the question
(i) Current is maximum at resonance, when $X_L=X_C$ : $\nu=\dfrac{1}{2\pi\sqrt{LC}}=\dfrac{1}{2\times3.14\times\sqrt{8\times2\times10^{-6}}}\approx39.80\text{ Hz}$ — this is called the (series) resonance frequency. (ii) At resonance $Z=R$ , so $I_0=\dfrac{E_0}{R}=\dfrac{200}{100}=2\text{ A}$ (iii) The current amplitude peaks sharply at $\omega=\omega_r$ for the smaller resistance $R_2$ (tall, narrow peak) and rises only modestly for the larger resistance $R_1$ (short, broad peak); the two curves coincide far from resonance. (iv) Sharpness of resonance is the ratio of the voltage developed across $L$ (or $C$ ) at resonance to the voltage developed across $R$ : $Q=\dfrac1R\sqrt{\dfrac LC}$ (equivalently $Q=\dfrac{\omega_r}{2\Delta\omega}$ ). A circuit becomes more selective (sharper resonance) when its resistance $R$ is small, i.e. when $Q$ is large.
20.↑ back to the question
(a) With $V_R=IR$ in phase with $I$ , $V_L=IX_L$ leading $I$ by $\pi/2$ , and $V_C=IX_C$ lagging $I$ by $\pi/2$ , the phasor sum gives $V=\sqrt{V_R^2+(V_L-V_C)^2}=I\sqrt{R^2+(X_L-X_C)^2}$ . So the current amplitude is $I_0=\dfrac{V_0}{\sqrt{R^2+\left(\omega L-\frac1{\omega C}\right)^2}}$ , and the phase angle is $\phi=\tan^{-1}\left(\dfrac{X_L-X_C}{R}\right)$ . (b) Resonance occurs when the current amplitude is maximum, i.e. when $\omega L=\dfrac1{\omega C} \Rightarrow \omega_0=\dfrac1{\sqrt{LC}}$ ; at resonance $Z=R$ (minimum impedance). Graph: current amplitude versus frequency peaks at $\omega=\omega_0$ for both resistances; the peak for the smaller resistance $R_2$ is taller and narrower (sharper), while for $R_1$ it is shorter and broader. Quality factor: $Q=\dfrac1R\sqrt{\dfrac LC}=\dfrac{\omega_0L}{R}=\dfrac{\omega_0}{2\Delta\omega}$ . A higher $Q$ gives a sharper resonance curve, so the circuit responds strongly only to a narrow band of frequencies near $\omega_0$ — this is what makes tuning (frequency selection) more precise.
21.Answer: (A)↑ back to the question
At resonance $X_L=X_C$ , so the voltage drops across inductor and capacitor are equal — options (c) and (d) are false. $Z=\sqrt{R^2+(X_L-X_C)^2}$ is large at very low and very high frequency and dips to its minimum value $R$ at resonance — i.e. as frequency is raised from a low value, $Z$ first decreases (towards resonance) and then increases again. Hence option (a).
22.Answer: (A)↑ back to the question
$X_L=\omega L$ , $X_C=\dfrac{1}{\omega C}$ $\dfrac{X_L}{X_C}=\dfrac{\omega L\times\omega C}{1}=\omega^2LC$ Hence option (a).
23.Answer: (C)↑ back to the question
A phase difference of $\pi/2$ between voltage and current is possible only when $L$ or $C$ (a purely reactive element) is present without any resistance limiting the phase angle to less than $\pi/2$ . If a resistor $R$ is present in the circuit (as in options combining $L$ and $R$ ), the phase angle $\phi=\tan^{-1}(X_L/R)$ is always strictly less than $\pi/2$ — it can never reach $\pi/2$ exactly. So $L$ and $R$ together cannot give $\phi=\pi/2$ . Hence option (c).
24.Answer: (D)↑ back to the question
For a series LCR circuit, the resistive and reactive voltage drops combine as phasors (perpendicular components), giving $Z=\sqrt{R^2+(X_L-X_C)^2}$ Hence option (d).
25.Answer: (C)↑ back to the question
$V_R=20$ V, $V_L=15$ V, $V_C=30$ V $V=\sqrt{V_R^2+(V_L-V_C)^2}=\sqrt{20^2+(15-30)^2}=\sqrt{400+225}=\sqrt{625}=25\text{ V}$ Hence option (c).
26.Answer: (A)↑ back to the question
In an ac circuit containing $L$ , $C$ and $R$ , the impedance $Z=\sqrt{R^2+(X_L-X_C)^2}$ is large away from resonance and falls to a minimum $Z=R$ at resonance, so the current $I=V/Z$ first increases, peaks at resonance, and then decreases as frequency is raised further. This behaviour requires the presence of both $L$ and $C$ together with $R$ (a pure $LC$ or $RC$ / $RL$ circuit does not show this rise-then-fall resonance behaviour). Hence option (a).
27.Answer: (B)↑ back to the question
Voltage and current in phase $\Rightarrow$ resonance, i.e. $X_L=X_C$ : $2\pi fL=\dfrac{1}{2\pi fC}$ $C=\dfrac{1}{4\pi^2f^2L}=\dfrac{1}{4\times(3.14)^2\times50^2\times80\times10^{-3}}=1.27\times10^{-4}\text{ F}=127\ \mu\text{F}$ Hence option (b).
28.Answer: (B)↑ back to the question
$X_C=\dfrac{1}{2\pi\nu C}=\dfrac{1}{2\pi\times50\times\dfrac{25}{\pi}\times10^{-6}}=400\ \Omega$ $Z=\sqrt{R^2+X_C^2}=\sqrt{300^2+400^2}=500\ \Omega$ $I=\dfrac{V}{Z}=\dfrac{200}{500}=0.4\text{ A}$ Hence option (b).
29.Answer: (C)↑ back to the question
Selectivity depends on the quality factor, $Q=\dfrac{\omega_0L}{R}$ . $Q$ (and hence selectivity) is large when $L$ is large and $R$ is small. Hence option (c).